Submersible Engine Design

[markdansie]

Hi Tommy
I am always in full admiration of your projects so well done.
However they are correct that what you have here will not work and has been tried many times.
If you email me markdansiebigpond,com or Skype me mark dansie (nsw austraia) I will show you two ways that should work. One is a maybe but it uses gravity to get the air in under presure the other will work and when I explain it you will see why (uses existing technology that will create a vacuum and suck the air in.
There is a third way.
I never got around to building them but your welcome to te information as I think your a decent person
Mark

[TommeyLReed]

This is what I have come up with, and how much energy is needed to make over unity.

If you only pump down at a depth of 10ft, .5(864cu/in)cu/ft
At 1 cu/ft of displacement of water is about 62.3lb at 1ft at depth, does this force not increase at a depth of 10 ft?
Pumping .5 cu/ft at a depth of 10 ft will increase the displacement each foot as it rise to the surfice, this will this also increase on lift too?
Its all theory, this is why it really needs to be look at.
Remember this 4.4 psi compare to a higher pressure in more efficient too make then compressing 90psi
Thanks you all the comment, now its almost ready to do real testing. This is where theory's become facts.
If i'm wrong so be it, but a prototype is better then any theory's!

Tommey Reed

[exnihiloest]

exnihiloest

i think you are a bit confussed.  the air expanding inside the cone does make a difference, if the cone only has 1 cubic foot of air to start.  at 10 feet it will look like only 83% of the full size.  thus the lift will be less.  you would have to use the average (91.5%) over the distance to figure how much work it can do.

if the cone is in a tube and sealed so water can't pass by, then your statement....

Do the maths.

Let V(h) the air volume of a cone, depending on the pressure. The variation of pressure is inversely proportional to the variation of height, thus we can write:
V(h)=V0+(V1-V0)*h/H
where h is any height where the cone is, V(h=0) = V0 is the inital volume at the bottom, V(h=H) = V1 is the volume near the surface at height H.

At height h, the mass of the volume of displaced water is:
M(h) = V(h)*d where d is the water volumetric mass (d=1 kg/dm³).
Its weight is P = M(h)*g = (V0+(V1-V0)*h/H)*d*g

The work of the Achimedes force is thus:
W = int[P*dh]  to be integrated from h=0 to h=H
W = int[(V0+(V1-V0)*h/H)*d*g*dh] = g*d* int[(V0+(V1-V0)*h/H)*dh]

int[h*dh] being 1/2*h², the calculus is straightforward:
W = g*d*(V0*H + 1/2*H*(V1-V0)) = g*d*(1/2*V0 + 1/2*V1)*H

This result is obvious because the volume is proportional to the pressure thus to the height, it follows that we can replace the variable pressure and the variable volume by a constant pressure and a constant volume equal to the mean values of the variable case. We can operate at constant pressure with a constant volume V= 1/2*V0 + 1/2*V1 and get strictly the same result.

This is explained by the Archimedes force and the weight of displaced water which are always balanced. Therefore the problem reduces to a pure problem of potential energy. The energy to compress air in the cone at the bottom, being the only energy to be considered as potential energy that can be recovered when the cone rises up, without extra gain.

[TommeyLReed]

This is two basic system.
One using gravity and the other using displacement.

http://www.youtube.com/watch?v=rn0yt0mylW8

http://www.youtube.com/watch?v=jC2k-mKsfEQ

Tom.

[powercat]

Excellent videos